Integration of dx/sin x cos^3x

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Integration of dx/sin x cos^3x

Let $$I = \int \sin x\, \cos 3x\,dx.$$ Use integration by parts: $$I = - \cos x\,\cos 3x - \int [-\cos x] [-3 \sin 3x]\, dx.$$ Pulling constants out: $$I = - \cos x\,\cos 3x - 3\int \cos x\, \sin 3x\, dx.$$ Using integration by parts again (and in the same way, so you do not just reverse what you did): $$I = - \cos x\,\cos 3x - 3\left[\sin x\, \sin 3x - \int \sin x\, [3 \cos 3x]\, dx \right].$$ Pulling constants out: $$I = - \cos x\,\cos 3x - 3\sin x\, \sin 3x + 9 \int \sin x\, \cos 3x\, dx.$$ But the last integral is $I$, possibly with a different constant of integration. $$I = - \cos x\,\cos 3x - 3\sin x\, \sin 3x + 9 I + C.$$ Shuffling around terms, $$8 I = \cos x\,\cos 3x + 3\sin x\, \sin 3x + C.$$ $$I = \frac{1}{8}\cos x\,\cos 3x + \frac{3}{8}\sin x\, \sin 3x + C.$$ To verify this, differentiate $I$. $$\frac{dI}{dx} = -\frac{1}{8}\sin x\,\cos 3x - \frac{3}{8}\cos x\,\sin 3x + \frac{3}{8}\cos x\, \sin 3x + \frac{9}{8}\sin x\, \cos 3x.$$ $$\frac{dI}{dx} = \sin x\,\cos 3x.$$ All right, so this is a boring subject; when I was teaching, this week tended to put my students to sleep. However, note that the definite integral from $0$ to $2\pi$ of this is $0$. In fact, choose any 2 of $\cos mx$ or $\sin nx$ with $0\le m$ and $1 \le n$. The definite integral will be $0$ unless you chose the same factor twice. And this is the start of Fourier Analysis. \[\int\sqrt{\sin x} \cos^3 x\ \text{ dx }\] \[ \text{ Let I} = \int\sqrt{\sin x} \cdot \cos^3 \text{ x dx }\]\[ = \int\sqrt{\sin x} \cdot \left( \cos^2 x \right) \cdot \text{ cos x dx }\]\[ = \int\sqrt{\sin x} \left( 1 - \sin^2 x \right) \cdot \text{ cos x dx}\]\[\text{ Putting sin x} = t\]\[ \Rightarrow \text{ cos x dx }= dt\]\[ \therefore I = \int\sqrt{t} \left( 1 - t^2 \right) \cdot dt\]\[ = \int t^\frac{1}{2} dt - \int t^\frac{1}{2} \cdot t^2 dt\]\[ = \int t^\frac{1}{2} dt \int t^\frac{5}{2} dt\]\[ = \frac{t^\frac{3}{2}}{\frac{3}{2}} - \frac{t^\frac{7}{2}}{\frac{7}{2}} + C\]\[ = \frac{2}{3} t^\frac{3}{2} - \frac{2}{7} t^\frac{7}{2} + C\]\[ = \frac{2}{3} \text{ sin }^\frac{3}{2} \text{ x }- \frac{2}{7} \text{ sin }^\frac{7}{2} \text{ x }+ C ..........\left[ \because t = \text{ sin x }\right]\] Concept: Indefinite Integral Problems Is there an error in this question or solution? Page 2 \[\int\frac{\sin 2x}{\sin^4 x + \cos^4 x} \text{ dx }\] \[\text{ Let I } = \int\frac{\sin 2x}{\sin^4 x + \cos^4 x}dx\]\[ = \int\frac{2 \text{ sin x }\cdot \text{ cos x dx}}{\sin^4 x + \cos^4 x}\]\[\text{Dividing numerator and denominator by} \cos^4 x\]\[ \Rightarrow \int\frac{2 \frac{\text{ sin x }\cdot \text{ cos x}}{\cos^4 x}dx}{1 + \tan^4 x}\]\[ \Rightarrow \int\frac{2 \tan x \cdot \text{ sec}^2 x dx}{1 + \left( \tan^2 x \right)^2}\]\[\text{ Putting tan}^2 x = t\]\[ \Rightarrow 2 \tan x \cdot \text{ sec}^2 \text{ x dx}\]\[ \therefore I = \int\frac{dt}{1 + t^2}\]\[ = \tan^{- 1} t + C\]\[ = \tan^{- 1} \left( \text{ tan}^2 x \right) + C......... \left[ \because t = \tan {}^2 x \right]\] Concept: Indefinite Integral Problems Is there an error in this question or solution? Page 3\[\text{ Let I }= \int\frac{1}{\sqrt{x^2 - a^2}} \text{ dx}\] \[\text{ Putting x = a sec } \] \[ \Rightarrow \text{ dx = a sec tan \text{ d}} \] \[ \therefore I = \int\frac{a \sec\theta \tan \text{ d} }{\sqrt{a^2 \sec^2 \theta - a^2}}\] \[ = \int\frac{{a \sec\theta\tan \text{ d} }}{a \cdot \tan\theta}\] \[ = \int\sec\tan \text{ d} \] \[ = \text{ ln }\left| \sec\theta + \tan\theta \right| + C\] \[ = \text{ ln} \left| \sec\theta + \sqrt{\sec^2 \theta - 1} \right| + C\] \[ = \text{ ln }\left| \frac{x}{a} + \sqrt{\left( \frac{x}{a} \right)^2 - 1} \right| + C\] \[ = \text{ ln} \left| \frac{x + \sqrt{x^2 - a^2}}{a} \right| + C\] \[ = \text{ ln} \left| x + \sqrt{x^2 - a^2} \right| - \text{ ln a} + C\] \[ = \text{ ln} \left| x + \sqrt{x^2 - a^2} \right| + C'\] \[\text{ where C' = C }- \text{ ln a }\]Page 4\[\text{ Let I } = \int\frac{dx}{\sqrt{x^2 - a^2}}\] \[\text{ Putting x} = a \tan \theta\] \[ \Rightarrow dx = a \sec^2 \text{ d }\] \ [ \therefore I = \int\frac{a \cdot se c^2\text{ d }}{\sqrt{a^2 \tan^2 \theta + a^2}}\] \[ = \int\frac{a \sec^2 \theta \cdot d\theta}{a\sqrt{1 + \tan^2 \theta}}\] \[ = \int\frac{\sec^2 \theta \cdot \text{ d }}{\sec\theta}\] \[ = \int\sec\theta \cdot d\theta\] \[ = \int\sec\theta \cdot d\theta\] \[ = \text{ ln } \left| \sec\theta + \tan\theta \right| + C\] \[ = \text{ ln }\left| \sec\theta + \sqrt{\sec^2 \theta - 1} \right| + C\] \[ = \text{ ln }\left| \frac{x}{a} + \sqrt{\frac{x^2}{a^2} - 1} \right| + C\] \[ = \text{ ln} \left| x + \sqrt{x^2 - a^2} \right| - \ln a + C\] \[ = \text{ ln }\left| x + \sqrt{x^2 - a^2} \right| + C'\] \[\text{ where C' = C - ln a }\]Page 5 \[\int\frac{1}{4 x^2 + 4x + 5} dx\] \[\text{ Let I }= \int\frac{dx}{4 x^2 + 4x + 1 + 4}\]\[ = \int\frac{dx}{\left( 2x \right)^2 + 2 \times 2x + 1 + 22}\]\[ = \int\frac{dx}{\left( 2x + 1 \right)^2 + 2^2}\]\[\text{ Putting }\left( 2x + 1 \right) = t\]\[ \Rightarrow 2 \text{ dx = dt }\]\[ \Rightarrow dx = \frac{dt}{2}\]\[ \therefore I = \frac{1}{2}\int\frac{dt}{t^2 + 2^2}\]\[ = \frac{1}{2} \times \frac{1}{2} \text{ tan}^{- 1} \left( \frac{t}{2} \right) + C\]\[ = \frac{1}{4} \text{ tan}^{- 1} \left( \frac{2x + 1}{2} \right) + C ....................\left[ \because t = \left( 2x + 1 \right) \right]\] Concept: Indefinite Integral Problems Is there an error in this question or solution? Page 6 \[\int\frac{1}{x^2 + 4x - 5} \text{ dx }\] \[\int\frac{1}{x^2 + 4x - 5}dx\]\[ = \int\frac{1}{x^2 + 4x + 4 - 4 - 5}dx\]\[ = \int\frac{1}{x^2 + 4x + 4 - 3^2}dx\]\[ = \int\frac{1}{\left( x + 2 \right)^2 - 3^2}dx\]\[ = \frac{1}{2 \times 3} \text{ ln} \left| \frac{x + 2 - 3}{x + 2 + 3} \right| + C ................. \left[ \because \int\frac{1}{x^2 - a^2}dx = \frac{1}{2a}\text{ ln }\left| \frac{x - a}{x + a} \right| + C \right]\]\[ = \frac{1}{6} \text{ ln } \left| \frac{x - 1}{x + 5} \right| + C\] Concept: Indefinite Integral Problems Is there an error in this question or solution? Page 7\[\text{ We have,} \]\[I = \int\frac{1}{1 - x - 4 x^2}dx\]\[ = \frac{1}{4}\int\frac{1}{\frac{1}{4} - - x^2 \frac{x}{4}}dx\]\[ = \frac{1}{4}\int\frac{1}{\frac{1}{4} - \left( x^2 + \frac{x}{4} \right)}dx\]\[ = \frac{1}{4}\int\frac{1}{\frac{1}{4} - \left\{ x^2 + + \left( \frac{1}{8} \right)^2 - \left( \frac{1}{8} \right)^2 \frac{x}{4} \right\}}dx\]\[ = \frac{1}{4}\int\frac{1}{\frac{1}{4} - \left( x + \frac{1}{8} \right)^2 + \frac{1}{64}}dx\]\[ = \frac{1}{4}\int\frac{1}{\frac{1}{4} + - \left( x + \frac{1}{8} \right)^2 \frac{1}{64}}dx\]\[ = \frac{1} {4}\int\frac{1}{\frac{16 + 1}{64} - \left( x + \frac{1}{8} \right)^2}dx\] \[ = \frac{1}{4}\int\frac{1}{\left( \frac{\sqrt{17}}{8} \right)^2 - \left( x + \frac{1}{8} \right)^2}dx\]\[ = \frac{1}{4} \times \frac{1}{2 \times \frac{\sqrt{17}}{8}} \text{ ln }\left| \frac{\frac{\sqrt{17}}{8} + x + \frac{1}{8}}{\frac{\sqrt{17}}{8} - x - \frac{1}{8}} \right| + C .................\left[ \because \int\frac{1}{a^2 - x^2}dx = \frac{1}{2a}\text{ ln }\left| \frac{a + x}{a - x} \right| + C \right]\]\[ = \frac{1}{\sqrt{17}} \text{ ln }\left| \frac{\frac{\sqrt{17} + 1}{8} + x}{\frac{\sqrt{17} - 1}{8} - x} \right| + C\]Page 8 \[\int\frac{1}{3 x^2 + 13x - 10} \text{ dx }\] \[\int\frac{1}{3 x^2 + 13x - 10}dx\]\[ = \frac{1} {3}\int\frac{1}{x^2 + \frac{13}{3}x - \frac{10}{3}}dx\]\[ = \frac{1}{3}\int\frac{1}{x^2 + \frac{13 x}{3} + \left( \frac{13}{6} \right)^2 - \left( \frac{13}{6} \right)^2 - \frac{10}{3}}dx\]\[ = \frac{1}{3}\int\frac{1}{\left( x + \frac{13}{6} \right)^2 - \frac{169}{36} - \frac{10}{3}}dx\]\[ = \frac{1}{3}\int\frac{1}{\left( x + \frac{13}{6} \right)^2 - \frac{169 - 120}{36}}dx\]\[ = \frac{1}{3}\int\frac{1}{\left( x + \frac{13}{6} \right)^2 - \left( \frac{17}{6} \right)^2}dx\]\[ = \frac{1}{3} \times \frac{1}{2 \times \frac{17}{6}} \text{ ln } \left| \frac{x + \frac{13}{6} - \frac{17}{6}}{x + \frac{13}{6} + \frac{17}{6}} \right| .............\left[ \because \int\frac{1}{x^2 - a^2}dx = \frac{1} {2a}\text{ ln }\left| \frac{x - a}{x + a} \right| + C \right]\]\[ = \frac{1}{17} \text{ ln}\left| \frac{x - \frac{2}{3}}{x + 5} \right| + C\]\[ = \frac{1}{17} \text{ ln }\left| \frac{3x - 2}{3x + 15} \right| + C\] Concept: Indefinite Integral Problems Is there an error in this question or solution? Page 9 \[\int\frac{\sin x}{\sqrt{\cos^2 x - 2 \cos x - 3}} \text{ dx }\] \[\text{ Let I }= \int\frac{\sin x}{\sqrt{\cos^2 x - 2 \cos x - 3}}dx\]\[\text{ Putting cos x = t}\]\[ \Rightarrow - \text{ sin x dx }= dt\]\[ \Rightarrow \text{ sin x dx } = - dt\]\[ \therefore I = - \int\frac{dt}{\sqrt{t^2 - 2t - 3}}\]\[ = - \int\frac{dt}{\sqrt{t^2 - 2t + 1 - 4}}\]\[ = - \int\frac{dt}{\sqrt{\left( t - 1 \right)^2 - \left( 2 \right)^2}}\]\[ = \text{ ln }\left| t - 1 + \sqrt{\left( t - 1 \right)^2 - 4} \right| + C ..........................\left[ \because \int\frac{1}{\sqrt{x^2 - a^2}}dx = \text{ ln}\left| x + \sqrt{x^2 - a^2} \right| + C \right]\]\[ = - \text{ ln }\left| \left( \cos x - 1 \right) + \sqrt{\cos^2 x - 2 \cos x - 3} \right| + C.................... \left[ \because t = \cos x \right]\] Concept: Indefinite Integral Problems Is there an error in this question or solution? Page 10\[\text{ Let I }= \int\sqrt{\text{ cosec x} - 1} \text{ dx}\] \[ = \int\sqrt{\frac{1}{\sin x} - 1} \text{ dx }\] \[ = \int\frac{\sqrt{1 - \sin x}}{\sqrt{\sin x}} \text{ dx }\] \[ = \int\frac{\sqrt{\left( 1 - \sin x \right) \left( 1 + \sin x \right)}}{\sqrt{\sin x \left( 1 + \sin x \right)}}\text{ dx }\] \[ = \int\frac{\sqrt{1 - \sin^2 x}}{\sqrt{\sin^2 x + \sin x}}\text{ dx}\] \[ = \int\frac{\cos x}{\sqrt{\sin^2 x + \sin x}}\text{ dx }\] \[\text{ Putting sin x = t }\] \[ \Rightarrow \text{ cos x dx = dt}\] \[ \therefore I = \int\frac{dt}{\sqrt{t^2 + t}}\] \[ = \int\frac{dt}{\sqrt{t^2 + t + \left( \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2}}\] \[ = \int\frac{dt}{\sqrt{\left( t + \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2}}\] \[ = \text{ ln }\left| t + \frac{1}{2} + \sqrt{\left( t + \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2} \right| + C ..............\left[ \because \int\frac{1}{\sqrt{x^2 - a^2}}dx = \text{ ln }\left| x + \sqrt{x^2 - a^2} \right| + C \right]\] \[ = \text{ ln} \left| t + \frac{1} {2} + \sqrt{t^2 + t} \right| + C\] \[ = \text{ ln }\left| \left( \sin x + \frac{1}{2} \right) + \sqrt{\sin^2 x + \sin x} \right| + C ............\left[ \because t = \sin x \right]\]Page 11 \[\int\frac{1}{\sqrt{3 - 2x - x^2}} \text{ dx}\] \[\text{ Let I} = \int\frac{1}{\sqrt{3 - 2x - x^2}}dx\] \[ = \int\frac{1}{\sqrt{3 - \left( x^2 + 2x + 1 - 1 \right)}}dx\] \[ = \int\frac{1} {\sqrt{4 - \left( x + 1 \right)^2}}dx\] \[\text{ Putting} \left( x + 1 \right) = t\] \[ \Rightarrow dx = dt\] \[ \therefore I = \int\frac{dt}{\sqrt{2^2 - t^2}}\] \[ = \sin^{- 1} \left( \frac{t}{2} \right) + C .................\left[ \because \int \frac{1}{\sqrt{a^2 - x^2}}dx = \sin^{- 1} \frac{x}{a} + C \right]\] \[ = \sin^{- 1} \left( \frac{x + 1}{2} \right) + C .....................\left[ \because t = \left( x + 1 \right) \right]\] Concept: Indefinite Integral Problems Is there an error in this question or solution? Page 12\[\text{ Let I } = \int\frac{\left( x + 1 \right)}{x^2 + 4x + 5}dx\] \[\text{ and let} \left( x + 1 \right) = A\frac{d}{dx}\left( x^2 + 4x + 5 \right) + B\] \[ \Rightarrow x + 1 = A \left( 2x + 4 \right) + B\] \ [ \Rightarrow x + 1 = \left( 2A \right)x + 4A + B\] \[\text{Equating the coefficients of like terms}\] \[2A = 1\] \[ \Rightarrow A = \frac{1}{2}\] \[\text{ and }\ 4A + B = 1\] \[ \Rightarrow 4 \times \frac{1}{2} + B = 1\] \[ \Rightarrow B = - 1\] \[ \therefore \left( x + 1 \right) = \frac{1}{2} \left( 2x + 4 \right) - 1\] \[ \therefore I = \int\left[ \frac{\frac{1} {2}\left( 2x + 4 \right) - 1}{x^2 + 4x + 5} \right]dx\] \[ = \frac{1}{2}\int\frac{\left( 2x + 4 \right)}{x^2 + 4x + 5}dx - \int\frac{1}{x^2 + 4x + 5}dx\] \[\text{ Putting x}^2 + 4x + 5 = t\] \[ \Rightarrow \left( 2x + 4 \right) dx = dt\] \[ \therefore I = \frac{1}{2}\int\frac{1}{t}dt - \int\frac{1}{x^2 + 4x + 4 + 1}dx\] \[ = \frac{1}{2}\int\frac{dt}{t} \int\frac{1}{\left( x + 2 \right)^2 + 1^2}dx \] \[ = \frac{1}{2} \text{ ln } \left| t \right| - \tan^{- 1} \left( \frac{x + 2}{1} \right) + C............. \left[ \because \int\frac{1}{x^2 + a^2}dx = \frac{1}{a} \tan^{- 1} \frac{x}{a} + C \right]\] \[ = \frac{1}{2} \text{ ln }\left| x^2 + 4x + 5 \right| - \tan^{- 1} \left( x + 2 \right) + C ...................\left[ \because t = x^2 + 4x + 5 \right]\] Page 13\[\text{We have}, \] \[I = \int\left( \frac{5x + 7}{\sqrt{\left( x - 5 \right)\left( x - 4 \right)}} \right) dx\] \[ = \int\left( \frac{5x + 7}{\sqrt{x^2 - 9x + 20}} \right) dx\] \[\text{ Let 5x + 7 }= A \frac{d}{dx} \left( x^2 - 9x + 20 \right) + B\] \[ \Rightarrow 5x + 7 = A \left( 2x - 9 \right) + B\] \[\text{Equating Coefficients of like terms}\] \[2A = 5\] \[ \Rightarrow A = \frac{5}{2}\] \[\text{ And }\] \[ - 9A + B = 7\] \[ \Rightarrow - 9 \times \frac{5}{2} + B = 7\] \[ \Rightarrow B = 7 + \frac{45}{2}\] \[ \Rightarrow B = \frac{59}{2}\] \[ \therefore I = \int\left( \frac{\frac{5}{2} \left( 2x - 9 \right) + \frac{59}{2}}{\sqrt{x^2 - 9x + 20}} \right) dx\] \[ = \frac{5} {2}\int\frac{\left( 2x - 9 \right) dx}{\sqrt{x^2 - 9x + 20}} + \frac{59}{2}\int\frac{dx}{\sqrt{x^2 - 9x + 20}}\] \[\text{ Putting x}^2 - 9x + 20 = t\] \[ \Rightarrow \left( 2x - 9 \right) dx = dt\] \[I = \frac{5}{2}\int\frac{dt}{\sqrt{t}} + \frac{59}{2}\int\frac{dx}{\sqrt{x^2 - 9x + \left( \frac{9}{2} \right)^2 - \left( \frac{9}{2} \right)^2 + 20}}\] \[ = \frac{5}{2}\int t^{- \frac{1}{2}} \text{ dt }+ \frac{59}{2}\int\frac{dx}{\sqrt{\left( x - \frac{9}{2} \right)^2 - \frac{81 + 80}{4}}}\] \[ = \frac{5}{2} \left[ \frac{t^{- \frac{1}{2} + 1}}{- \frac{1}{2} + 1} \right] + \frac{59}{2} \int\frac{dx}{\sqrt{\left( x - \frac{9}{2} \right)^2 - \left( \frac{1}{2} \right)^2}}\] \[ = \frac{5}{2} \times 2\sqrt{t} + \frac{59}{2} \text{ log }\left| \left( x - \frac{9}{2} \right) + \sqrt{\left( x - \frac{9}{2} \right)^2 - \left( \frac{1}{2} \right)^2} \right| + C\] \[ = 5\sqrt{t} + \frac{59}{2} \text{ log} \left| \left( x - \frac{9}{2} \right) + \sqrt{x^2 - 9x + 20} \right| + C\] \[ = 5\sqrt{x^2 - 9x + 20} + \frac{59}{2} \text{ log }\left| \left( x - \frac{9}{2} \right) + \sqrt{x^2 - 9x + 20} \right| + C\]Page 14\[\text{ Let I }= \int\sqrt{\frac{1 + x}{x}}dx\] \[ = \int\frac{\sqrt{1 + x}}{\sqrt{x}} \times \frac{\sqrt{1 + x}}{\sqrt{1 + x}}dx\] \[ = \int\left( \frac{1 + x}{\sqrt{x^2 + x}} \right)dx\] \[\text{ Let x }+ 1 = A\frac{d}{dx}\left( x^2 + x \right) + B\] \[ \Rightarrow x + 1 = A \left( 2x + 1 \right) + B\] \[ \Rightarrow x + 1 = \left( 2A \right)x + A + B\] \[\text{Equating the coefficients of like terms}\] \[2A = 1\] \[ \Rightarrow A = \frac{1}{2}\] \[\text{ and A + B = 1 }\] \[ \Rightarrow \frac{1}{2} + B = 1\] \[ \therefore B = \frac{1}{2}\] \[ \therefore I = \int\frac{\left( x + 1 \right)}{\sqrt{x^2 + x}}dx\] \[ = \int\left( \frac{\frac{1}{2} \left( 2x + 1 \right) + \frac{1}{2}} {\sqrt{x^2 + x}} \right)dx\] \[ = \frac{1}{2}\int\frac{\left( 2x + 1 \right)}{\sqrt{x^2 + x}}dx + \frac{1}{2}\int\frac{1}{\sqrt{x^2 + x}}dx\] \[\text{ Putting x}^2 + x = t\] \[ \Rightarrow \left( 2x + 1 \right) dx = dt\] \[ \therefore I = \frac{1}{2}\int\frac{1}{\sqrt{t}}dt + \frac{1}{2}\int\frac{1}{\sqrt{x^2 + x + \left( \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2}}dx\] \[ = \frac{1}{2}\int\frac{1}{\sqrt{t}}dt + \frac{1}{2}\int\frac{1}{\sqrt{\left( x + \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2}}dx\] \[ = \frac{1}{2}\int t^{- \frac{1}{2}} dt + \frac{1}{2}\int\frac{1}{\sqrt{\left( x + \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2}}dx\] \[ = \frac{1}{2} \times 2 \sqrt{t} + \frac{1}{2} \text{ ln }\left| x + \frac{1}{2} + \sqrt{\left( x + \frac{1}{2} \right)^2 - \frac{1}{4}} \right| + C............ \left[ \because \int\frac{1}{\sqrt{x^2 - a^2}}dx = \text{ ln }\left| x + \sqrt{x^2 - a^2} \right| + C \right]\] \[ = \sqrt{t} + \frac{1}{2} \text{ ln} \left| x + \frac{1}{2} + \sqrt{x^2 + x} \right| + C\] \[ = \sqrt{x^2 + x} + \frac{1}{2} \text{ ln} \left| \left( x + \frac{1}{2} \right) + \sqrt{x^2 + x} \right| + C................... \left[ \because t = x^2 + x \right]\]Page 15\[\text{ Let I } = \int\frac{\sqrt{1 - x}}{\sqrt{x}}dx\]\[ = \int\left( \frac{\sqrt{1 - x} \cdot \sqrt{1 - x}}{\sqrt{x} \cdot \sqrt{1 - x}} \right) dx\]\[ = \int\frac{\left( 1 - x \right)}{\sqrt{x - x^2}}dx\]\[\text{ Let} \left( 1 - x \right) = A\frac{d}{dx}\left( x - x^2 \right) + B\]\[ \Rightarrow 1 - x = A \left( 1 - 2x \right) + B\]\[ \Rightarrow 1 - x = - \left( 2A \right) x + A + B\]\[\text{Equating coefficients of like terms}\]\[ - 2A = - 1\]\[ \Rightarrow A = \frac{1}{2}\]\[\text{ and A + B = 1 }\]\[ \Rightarrow \frac{1}{2} + B = 1\]\[ \therefore B = \frac{1}{2}\]\[ \therefore I = \int\frac{\frac{1}{2} \left( 1 - 2x \right) + \frac{1}{2}}{\sqrt{x - x^2}}dx\]\[ = \frac{1}{2}\int\frac{\left( 1 - 2x \right)}{\sqrt{x - x^2}}dx + \frac{1}{2}\int\frac{1}{\sqrt{x - x^2 + \left( \frac{1}{2} \right)^2 - \left( \frac{1}{2} \right)^2}}dx\]\[ = \frac{1}{2}\int\frac{\left( 1 - 2x \right)}{\sqrt{x - x^2}}dx + \frac{1}{2}\int\frac{1}{\sqrt{\left( \frac{1}{2} \right)^2 - \left( x^2 - x + \frac{1}{2^2} \right)}}dx\]\[ = \frac{1}{2}\int\frac{\left( 1 - 2x \right)}{\sqrt{x - x^2}}dx + \frac{1}{2}\int\frac{1}{\sqrt{\left( \frac{1}{2} \right)^2 - \left( x - \frac{1}{2} \right)^2}}dx\] \[\text{ Putting x - x}^2 =\text{ t in the first integral }\] \[ \Rightarrow \left( 1 - 2x \right)\text{ dx } = dt\] \[ \therefore I = \frac{1}{2}\int\frac{1}{\sqrt{t}}dt + \frac{1}{2}\int\frac{1}{\sqrt{\left( \frac{1}{2} \right)^2 - \left( x - \frac{1}{2} \right)^2}}dx\] \[ = \frac{1}{2}\int t^{- \frac{1}{2}} dt + \frac{1}{2}\int\frac{dx}{\sqrt{\left( \frac{1}{2} \right)^2 - \left( x - \frac{1}{2} \right)^2}}\] \[ = \frac{1}{2} \times 2\text{ t}^\frac{1}{2} + \frac{1}{2} \times \sin^{- 1} \left( \frac{x - \frac{1}{2}}{\frac{1}{2}} \right) + C................ \left[ \because \int\frac{1}{\sqrt{a^2 - x^2}}dx = \sin^{- 1} \frac{x}{a} + C \right]\] \[ = \sqrt{t} + \frac{1}{2} \text{ sin}^{- 1} \left( 2x - 1 \right) + C\] \[ = \sqrt{x - x^2} + \frac{1}{2} \text{ sin}^{- 1} \left( 2x - 1 \right) + C ..................\left[ \because t = x - x^2 \right]\]Page 16\[\text{ We have,} \] \[I = \int\frac{\sqrt{a} - \sqrt{x}}{1 - \sqrt{ax}} \text{ dx }\] \[I = \frac{1}{\sqrt{a}}\int\frac{1 + a - 1 - \sqrt{ax}}{1 - \sqrt{ax}} \text{ dx }\] \[I = \frac{1}{\sqrt{a}}\int\frac{1 - \sqrt{ax}}{1 - \sqrt{ax}} dx + \frac{1}{\sqrt{a}}\int\frac{a - 1}{1 - \sqrt{ax}} \text{ dx }\] \[I = \frac{1}{\sqrt{a}}\int dx + \frac{a - 1}{\sqrt{a}}\int\frac{1}{1 - \sqrt{ax}} \text{ dx }\] \[I = \frac{1}{\sqrt{a}}x + \frac{a - 1}{\sqrt{a}}\int\frac{1}{1 - \sqrt{ax}} \text{ dx}\] \[\text{ Let,} \] \[ I_1 = \int\frac{1}{1 - \sqrt{ax}} \text{ dx }\] \[\text{ Put ax = z}^2 \] \[ \Rightarrow adx = \text{ 2 }zdz\] \[ I_1 = \frac{1}{a}\int\frac{2z}{1 - z}\text{ dz}\] \[ I_1 = \frac{1} {a}\int\frac{2z - 2 + 2}{1 - z} \text{ dz }\] \[ I_1 = \frac{1}{a}\int\frac{2z - 2}{1 - z} \text{ dz } + \frac{1}{a}\int\frac{2}{1 - z} \text{ dz }\] \[ I_1 = \frac{- 2}{a}\int\frac{1 - z}{1 - z} \text{ dz } + \frac{1}{a}\int\frac{2}{1 - z} \text{ dz }\] \[ I_1 = \frac{- 2}{a}\int \text{ dz } + \frac{1}{a}\int\frac{2}{1 - z} \text{ dz }\] \[ I_1 = \frac{- 2}{a}z \frac{2}{a}\text{ log }\left| 1 - z \right| + C_1 \] \[ I_1 = \frac{- 2\sqrt{ax}}{a} - \frac{2}{a}\text{ log}\left| 1 - \sqrt{ax} \right| + C_1 \] \[I = \frac{1}{\sqrt{a}}x + \frac{a - 1}{\sqrt{a}}\left( \frac{- 2\sqrt{ax}}{a} - \frac{2}{a}\text{ log }\left| 1 - \sqrt{ax} \right| \right) + C\] Note: The answer in indefinite integration may vary depending on the integral constant.Page 17 \[\int\frac{1}{\left( \sin x - 2 \cos x \right) \left( 2 \sin x + \cos x \right)} \text{ dx }\] \[\int\frac{1}{\left( \sin x - 2 \cos x \right) \left( 2 \sin x + \cos x \right)} dx\] Dividing numerator and denominator by cos2x we get , \[I = \int\frac{\frac{1}{\cos^2 x}}{\left( \tan x - 2 \right) \left( 2 \tan x + 1 \right)}dx\]\[ = \int\frac{\sec^2 x}{\left( \tan x - 2 \right) \left( 2 \tan x + 1 \right)} dx\]\[\text{ Putting tan x = t }\]\[ \Rightarrow \text{ sec}^2 \text{ x dx} = dt\]\[ \therefore I = \int\frac{1}{\left( t - 2 \right) \left( 2t + 1 \right)}dt\]\[ = \int\frac{1}{2 t^2 + t - 4t - 2}dt\]\[ = \int\frac{1}{2 t^2 - 3t - 2}dt\]\[ = \frac{1}{2}\int\frac{1}{t^2 - \frac{3t}{2} - 1}dt\]\[ = \frac{1} {2}\int\frac{1}{t^2 - \frac{3}{2}t + \left( \frac{3}{4} \right)^2 - \left( \frac{3}{4} \right)^2 - 1}dt\]\[ = \frac{1}{2}\int\frac{1}{\left( t - \frac{3}{4} \right)^2 - \frac{9}{16} - 1}dt\]\[ = \frac{1}{2}\int\frac{1}{\left( t - \frac{3}{4} \right)^2 - \left( \frac{5}{4} \right)^2}dt\]\[ = \frac{1}{2} \times \frac{1}{2 \times \frac{5}{4}} \text{ ln }\left| \frac{t - \frac{3}{4} - \frac{5}{4}}{t - \frac{3}{4} + \frac{5}{4}} \right| + C ....................\left[ \because \int\frac{1}{x^2 - a^2}dx = \frac{1}{2a}\text{ ln }\left| \frac{x - a}{x + a} \right| + C \right]\]\[ = \frac{1}{5} \text{ ln }\left| \frac{t - 2}{t + \frac{1}{2}} \right| + C\]\[ = \frac{1}{5} \text{ ln } \left| \frac{2 \left( t - 2 \right)}{2t + 1} \right| + C\]\[ = \frac{1}{5} \text{ ln }\left| \frac{2 \left( \tan x - 2 \right)}{2 \tan x + 1} \right| + C................ \left[ \because t = \tan x \right]\]\[ = \frac{1}{5} \text{ ln} \left| \frac{\tan x - 2}{2 \tan x + 1} \right| + \frac{1}{5} \text{ ln 2 + C }\]\[ = \frac{1}{5} \text{ ln }\left| \frac{\tan x - 2}{2 \tan x + 1} \right| + C'\]\[\text{ where } \]\[C' = C + \frac{1} {5} \text{ ln 2 }\] Concept: Indefinite Integral Problems Is there an error in this question or solution? Page 18 \[\text{ Let I } = \int\frac{1}{4 \sin^2 x + 4 \sin x \cdot \cos x + 5 \cos^2 x}dx\] Dividing numerator and denominator by cos2x we get \[I = \int\frac{\sec^2 x}{4 \tan^2 x + 4 \tan x + 5}dx\] \[\text{ Putting tan x = t}\] \[ \Rightarrow \text{ sec}^2 \text{ x dx = dt }\] \[ \therefore I = \int\frac{dt}{4 t^2 + 4t + 5}\] \[ = \frac{1}{4}\int\frac{dt}{t^2 + t + \frac{5}{4}}\] \[ = \frac{1}{4}\int\frac{dt}{t^2 + t + \frac{1}{4} - \frac{1}{4} + \frac{5}{4}}\] \[ = \frac{1}{4}\int\frac{dt}{\left( t + \frac{1}{2} \right)^2 + 1^2}\] \[ = \frac{1}{4} \times \tan^{- 1} \left( t + \frac{1}{2} \right) + C.......... \left[ \because \int\frac{1}{x^2 + a^2}dx = \frac{1}{a} \tan^{- 1} \frac{x}{a} + C \right]\] \[ = \frac{1}{4} \tan^{- 1} \left( \frac{2t + 1}{2} \right) + C\] \[ = \frac{1}{4} \tan^{- 1} \left( \frac{2 \tan x + 1}{2} \right) + C...........\left[ \because t = \tan x \right]\] \[ = \frac{1}{4} \tan^{- 1} \left( \tan x + \frac{1}{2} \right) + C\]Page 19\ [\text{ Let I } = \int\frac{1}{a + b \tan x}dx\] \[ = \int\frac{1}{a + b \frac{\sin x}{\cos x}}dx\] \[ = \int\frac{\cos x \cdot}{a \cos x + b \sin x}dx\] \[\text{ Let } \cos x = \text{ A }\frac{d}{dx} \left( a \cos x + b \sin x \right) + \text{ B }\left( a \cos x + b \sin x \right)\] \[ \Rightarrow \cos x = A \left( - a \sin x + b \cos x \right) + B \left( a \cos x + b \sin x \right)\] \[1 \cdot \cos x = \left( Ab + B \cdot a \right) \cos x + \sin x\left( - A \cdot a + B \cdot b \right)\] \[\text{Equating coefficients of like terms}\] \[ A \cdot b + B \cdot a = 1 . . . \left( 1 \right)\] \[ - A \cdot a + B \cdot b = 0 . . . \left( 2 \right)\] \[\text{Multiplying equation} \left( 1 \right) \text{by a and eq} \left( 2 \right) \text{by b and then adding them} \] \[ A \cdot ab + B \cdot a^2 = a\] \[ - A \cdot a \cdot b + B b^2 = 0\] \[ \Rightarrow B = \frac{a}{a^2 + b^2}\] \[\text{Substituting the value of B in eq} \left( 1 \right)\] \[ \Rightarrow A \cdot b + \frac{a^2}{a^2 + b^2} = 1\] \[ \Rightarrow A \cdot b = 1 - \frac{a^2}{a^2 + b^2}\] \[ \Rightarrow A = \frac{b}{a^2 + b^2}\] \[ \therefore I = \frac{b}{a^2 + b^2}\int\left( \frac{- a \sin x + b \cos x}{a \cos x + b \sin x} \right)dx + \frac{a}{a^2 + b^2}\int\left( \frac{a \cos x + b \sin x}{a \cos x + b \sin x} \right)dx\] \[ = \frac{b}{a^2 + b^2}\int\left( \frac{- a \sin x + b \cos x}{a \cos x + b \sin x} \right)dx + \frac{a}{a^2 + b^2}\int dx\] \[\text{ Putting a cos x + b sin x = t in the Ist integral}\] \[ \Rightarrow \left( - a \sin x + b \cos x \right)dx = dt\] \[ \therefore I = \frac{b}{a^2 + b^2}\int\frac{dt}{t} + \frac{a}{a^2 + b^2}\int dx\] \[ = \frac{b}{a^2 + b^2} \text{ ln }\left| t \right| + \frac{ax}{a^2 + b^2} + C\] \[ = \frac{b}{a^2 + b^2} \text{ ln} \left| a \cos x + b \sin x \right| + \frac{ax}{a^2 + b^2} + C................ \left[ \because t = a \cos x + b \sin x \right]\]Page 20 \[\int\frac{1}{\sin^2 x + \sin 2x} \text{ dx }\] \[\text{ Let I } = \int\frac{1}{\sin^2 x + \sin 2x}dx\] \[ = \int\frac{1}{\sin^2 x + 2 \sin x \cdot \cos x}dx\] Dividing numerator and denominator by cos2x, we get \[I = \int\frac{\frac{1}{\cos^2 x}}{\tan^2 x + 2 \tan x}dx\]\[ = \int\frac{\sec^2 x}{\tan^2 x + 2 \tan x} dx\]\[\text{ Putting tan x = t}\]\[ \Rightarrow \text{ sec}^2 \text{ x dx = dt }\]\[ \therefore I = \int\frac{1}{t^2 + 2t}dt\]\[ = \int\frac{1}{t^2 + 2t + 1 - 1}dt\]\[ = \int\frac{1}{\left( t + 1 \right)^2 - 1^2}dt\]\[ = \frac{1}{2} \text{ ln} \left| \frac{t + 1 - 1}{t + 1 + 1} \right| + C\]\[ = \frac{1}{2} \text{ ln } \left| \frac{t}{t + 2} \right| + C \]\[ = \frac{1}{2} \text{ ln} \left| \frac{\tan x}{\tan x + 2} \right| + C ............\left[ \because t = \tan x \right]\] Concept: Indefinite Integral Problems Is there an error in this question or solution?

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